Leftmost Digit

Problem Description

Given a positive integer N, you should output the leftmost digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the leftmost digit of N^N.

 

Sample Input

234

 

Sample Output

22
Hint
In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.
 

 

Author

 

 

 

Ignatius.L

 

 

一看题上的 数值范围 就知道一般的数值类型没法表示 考虑到__int64

又因为n^n数值太大 考虑用指数表示法

 则有n^n=a*10^x;

等号两边对10取对数得 n*log10n=lga+x;

则lga=(n*lgn-x)

x可以通过对n*lgn向下取整求得

则有lga=(n*lgn-[n*lgn])

所以有以下代码：

#include<stdio.h>
#include<math.h>
int main()
{
 int t;
 scanf("%d",&t);
 while(t--)
 {
  __int64 n,s;
  double m;
  scanf("%I64d",&n);
  m=n*log10(n+0.0);
  m-=(__int64)m;
  s=(__int64)pow(10,m);
  printf("%I64d\n",s);
 }
 return 0;
}